TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-03

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A module homomorphism vanishing on $N$ factors uniquely through $M/N$

Statement

Let $f:M\to P$ be a module homomorphism and let $N\le M$ satisfy $N\subseteq\ker f$ . There is a unique module homomorphism

$\bar f:M/N\longrightarrow P$

such that $\bar f(m+N)=f(m)$ , equivalently $f=\bar f\circ\pi$ .

Facts & Assumptions

Given: A module homomorphism $f:M\to P$ and a submodule $N\le M$ with $N\subseteq\ker f$ .

[L1]

The canonical map $\pi:M\to M/N$ is a surjective module homomorphism with kernel $N$ (The canonical map $M\to M/N$ is a surjective module homomorphism with kernel $N$ ; thus every submodule is a kernel).

[L2]

A group homomorphism that kills a normal subgroup factors uniquely through the group quotient (A homomorphism that kills a normal subgroup factors uniquely through the quotient group).

[L3]

A module homomorphism is additive and scalar-preserving, and its kernel is the preimage of zero (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · constructive

1.1

By [L1], the canonical map $\pi:M\to M/N$ is an additive-group quotient map with kernel $N$ . Viewing the modules as additive groups, [L3] makes $f$ a group homomorphism and the hypothesis says it kills $N$ . By [L2], define the unique additive homomorphism $\bar f$ by $\bar f(m+N)=f(m)$ , with $f=\bar f\circ\pi$ .

L1L2L3givenconstruct

2.1

For every coset, $\bar f(r(m+N))=\bar f(rm+N)=f(rm)=rf(m)=r\bar f(m+N)$ , so $\bar f$ is scalar-preserving.

step 1.1L3given

3.1

Thus $\bar f$ is a module homomorphism with the required factorisation.

step 1.1step 2.1L3

4.1

Any module-homomorphism factor is in particular an additive-group factor, so the uniqueness in step 1.1 proves its uniqueness as a module homomorphism.

step 1.1L2L3discharge-construct: final∎

Depends on

Used by

First isomorphism theorem for modules: M/ker f congimf Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 30 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

McGerty, Algebra II: Rings and Modules, Section 3 (standard reference, not scraped)

A module homomorphism vanishing on NNN factors uniquely through M/NM/NM/N